Don't Ask Marilyn - Part 4 or Let's Do Math!

I want to conclude this series on the Marilyn von Savant's die roll problem (see the original column from Parade magazine and my part 1, part 2, and part 3) with a proper probability calculation, even though my analysis using entropy is spot on.  When I'm done, I will be writing about beer.  Here's a brief highlight of what's  to come.  Beer bubbles are cool, and drinking beer may aid women in preventing osteoporosis.

My reader thinks that Marilyn's die roll problem is one of conditional probability.  I disagree.

Conditional probability means "what is the probability that an event occurs (let's call this event B) if we already know that another event (let's call this event A) has already occurred."  To use the phrase of my reader and one that is used in conditional probability, "what is the probability that event B occurs given that event A has already occurred."

Two examples to illustrate:

• I throw a die, and I roll a 4.  What's the probability that the next roll is a 1 given that the first roll is a 4?  Since these two events are independent, the probability of rolling a 1 is 1/6.  In other words, knowing I threw a 4 does not affect the next roll.
• I throw a die, but don't tell you what I roll, except I do tell you it's not a 5 or a 6.  What's the probability that it's a 4 given that it's not a 5 or a 6?  The answer - 1/4.

MATH ALERT!   In chapter 4 of "Introduction to Probability" by Grinstead and Snell, conditional probability is calculated with the formula

In order to calculate the conditional probability P(B|A) [the probability that B occurs given that A has already happened], we need to calculate P(A) [the probability that A happened] and P(A and B) [the probability of A and B; the upside down U is the mathematical symbol for union and can be understood to mean 'and.']

Let's examine the two examples from above.

• P(A) = 1/6.  P(A and B) = 1/36.  Then P(B|A) = 1/6.
• P(A) = 4/6, since knowing that the roll is not 5 or 6 is the same as knowing that it is a 1, 2, 3, or 4.  P(A and B) = 1/6.  Then P(B|A) = 1/4.

Now I examine Marilyn's problem.  There are two possible outcomes. Roll (a) 11111111111111111111 and roll (b) 66234441536125563152.  

So let's assign events and be careful.  Event B is roll (b), since I'm interested in knowing what the probability of rolling (b) given that a die has been rolled 20 times.  Then event A is rolling a die 20 times.

P(A and B) = the probability of rolling (b) and rolling a die 20 times = 2.7 x 10^-16.

P(A) = the probability of rolling a die 20 times = 1.

Therefore, Then P(B|A) = the probability of rolling (b) given that a die has been rolled 20 times = 2.7 x 10^-16.

See.  George Alland, the math teacher that challenged Marilyn, is correct.  Marilyn is wrong in insisting "It was far more likely to have been that mix than a series of ones."  Marilyn corrects her mistake when she writes that "a jumble of numbers" is more likely.  This was my point in part 1, part 2, and part 3.

There is one more issue I'd like to address. Something that might help clarify the difference between 'that mix' and 'a jumble.'  There is in math the concepts of permutations and combinations.

A permutation is a arrangement of things in which order of the things matters.  Suppose you buy a raffle ticket an your number is 407.  Do you win if the number called is 074?  No, because the order of those digits matter.  Now suppose you buy a lottery ticket. and your numbers are 1, 13, 25, 26, 33, and 42.  Do you win when you see the ping pong balls come up 13, 26, 42, 33, 1, 25?  Yes you do, because the only thing that matters is the combination of numbers, not the order in which they are drawn.  When Marilyn wrote 'that mix' she - perhaps inadvertently - specified a particular order, one permutation.  When she wrote 'a jumble' - perhaps she caught her previous mistake - she now highlights the combination, not the order.

UPDATE:
How does the probability change when when we consider the combination rather than the permutation?

Event B is roll (b), since I'm interested in knowing what the probability of rolling (b) given that either (a) or (b) is rolled. Event A is now rolling (a) or (b).

P(A and B) = 2.7 x 10^-16.
P(A) = 5.4 x 10^-16.
Therefore, Then P(B|A) = 1/2.

But if I change the event B to rolling three 1s, three 2s, three 3s, three 4s, four 5s, and four 6s,  then

P(A and B) = 0.239.

P(A) = 0.239+ 2.7 x 10^-16.

Therefore, Then P(B|A) = 1.

I hope this settles the matter. I need a beer.

p.s. Many thanks to my reader. He truly highlights the need for all of us to be clear in our writing and our mathematics. I hope that all my readers hold me to such standards.